Write the standard equation for the circle center (–6, –8), that passes through (0, 0). How would someone do this? I have been looking through all my notes from class and I don't understand how to do it.
very easy
You should be familiar with the standard form that shows the centre.
the equation would be
(x+6)^2 + (y+8)^2 = r^2
but (0,0) lies on it, so
6^2 + 8^2= r^2 = 100
(x+6)^2 + (y+8)^2 = 100
Thankyou very much :)
To find the equation of a circle with a given center and a point that lies on the circle, you can use the equation for a circle in standard form, which is:
(x - h)^2 + (y - k)^2 = r^2
where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle.
In this case, the center of the circle is (-6, -8), and the point that lies on the circle is (0, 0).
To find the radius, you can use the distance formula between the center point and the point that lies on the circle. The distance formula is:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Using the coordinates of the center (-6, -8) and the given point (0, 0), the distance formula can be used as follows:
r = sqrt((-6 - 0)^2 + (-8 - 0)^2)
Simplifying this expression:
r = sqrt(36 + 64)
r = sqrt(100)
r = 10
Now that you have the center (-6, -8) and the radius (10), you can substitute these values into the equation for a circle in standard form:
(x - (-6))^2 + (y - (-8))^2 = 10^2
(x + 6)^2 + (y + 8)^2 = 100
So, the standard equation for the circle with a center at (-6, -8) that passes through (0, 0) is (x + 6)^2 + (y + 8)^2 = 100.